A particle is projected from the bottom of an inclined plane of inclination $$300$$ with velocity of $$40$$ $$m/s$$ at an angle of $$600$$ with horizontal. Then (Take$$ $$g=10$$ $$m/s2$$ $$)

Question

A particle is projected from the bottom of an inclined plane of inclination  $$300$$ with velocity of $$40$$ $$m/s$$ at an angle of $$600$$ with horizontal. Then  (Take$$ $$g=10$$ $$m/s2$$ $$)

Infinity Learn · Accepted Answer

Velocity vector will be parallel to the plane when component of velocity Perpendicular to the plane is zero. Here  α$$=600$$,β$$=300$$  Velocity vector will be parallel to the plane when component of velocity Perpendicular to the plane is $$zeroVy=uy+ayt=0vy=usin$$⁡$$($$α−β$$)$$−$$(gcos$$⁡β$$)t=0$$   In this problem, β$$=30$$∘,α$$=60$$∘ and $$u=40msecvy=40$$×$$sin$$⁡$$30$$∘−$$10cos$$⁡$$30$$∘$$=0$$⇒$$20=53t$$⇒$$t=43sec$$ At this instant the particle will have vx component only $$vx=ux+axt$$  $$vx=40cos$$⁡$$30$$∘−$$10$$×$$sin$$⁡$$30$$∘×$$43vx=403m/$$$$sec$$ Speed of the particle  $$=403m/s$$

A particle is projected from the bottom of an inclined plane of inclination $3 0^{0}$ with velocity of 40 m/s at an angle of $60^{0}$ with horizontal. Then (Take $g = 10 m / s^{2}$ )

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A particle is projected from the bottom of an inclined plane of inclination 300 with velocity of 40 m/s at an angle of 600 with horizontal. Then (Take g=10 m/s2 )

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A particle is projected from the bottom of an inclined plane of inclination $3 0^{0}$ with velocity of 40 m/s at an angle of $60^{0}$ with horizontal. Then (Take $g = 10 m / s^{2}$ )